1. How much heat is needed to raise the temperature of the copper kettle and its contenets to 80 C?
m,water= .5kg
Ti=20 C
Tf=80 C
m, copper= 1kg
Cwater= 4.18j/kgC
Ccopper= 0.385kj/KgC

Delta T = Tf-Ti
= 80 - 20
= 60 C

Q = mcT

Qwater= (0.5)(4.18)(60)
Qwater= 125.4kj

Qcopper = (1)(0.385)(60)
Qcopper = 23.1kj

Qtotal= Qwater + Qcopper
Qtotal= 125.4 + 23.1
Qtotal= 148.5kj

The total heat required is 148.5kj.

2. The heat of neutralization of HCl with NaOH is 57 kJ/mol (for every mole of NaOH which is neutralized by the HCl, 57 kJ of heat is released).

a) How much heat is released by mixing 50 mL of 1.0 M acid with 50 mL of 1.0 M base?

50 mL NaOH/x mol (times) 1000ml NaOH/1 mol
(cross-multiply)
50 = 1000x
x = 0.05
x = 0.05 moles

1 mol NaOH/57 kJ (times) 0.05 mol NaOH/ x kJ
(cross-multiply)
x = (0.05)(57)
x = 2.85
x = 2.85 kJ

Therefore 2.8 kJ are released.

b) If both solutions were initially at 20 degrees Celsius, what will be the final temperature of the mixture?
q = mc∆T
q = 2.85kJ (x 1000)
= 2850 J
m = 50 mL x (1g / mL)
= 50 g (x2)
= 100 g
c = 4.18J/g C

2850 = (100) (4.18) ∆T
2850 = 418∆T
2850/418 = 418∆T/418
∆T = 6.82
∆T = 6.8 degrees Celsius
20 degrees Celsius (initial temp.) + 6.8 degrees Celsius (change in temp) = 26. 8 degrees Celsius

Therefore temperature of the mixture (or temp. of surroundings since you're referring to the water) is 26.8 degrees Celsius.

3. The burning of 1 mol of methane releases 900kJ. How much methane must be burned to heat 100L of water for a bath from a starting temperature of 20 C to a final temperature of 45 C?
c(H20)=4.18 J/g*C

q=?
q(CH4)=900 kJ/mol
n(CH4)=?
m(CH4)=?

V(H2O)=100 mL
m(H2O)=100 g

T(f)=45 C
T(i)=20 C
∆T=25 C


q=mc∆T
q=(100)(4.18)(25)
q=10 450 kJ

1 mol =........n.........
.900kJ........10 450kJ
n=11.61111...mol

m(CH4)=mm(CH4)*n
............=(16)(11.6111...)
............=185.78 g
Therefore, 185.78g of methane must be burnt to heat the bath water 25 C!


DON'T BOTHER FORMATTING WITH SPACES. IT JUST GETS MESSED UP. DX

4.
Givens
per gram of wax= 52.5 J of energy released (Q)
Density of water= 1g/ml
c of water= 4.18 J/Cg
D=m/v
1m3= 1kJ
Vwater =0.5x10^3 L= 0.5 g= m of water

Required
how many grams of wax required? and how much Q is needed?


Qsurr= 0.5g(4.18J/Cg)(20C)= 41.8 J


52.5/1g = 41.8J/x
x=( 41.8J(1g))/ 52.5 kJ
x= - 0.796g/ J
x= 796g
Solution: The mass of wax required to raise the temp. of the water by 20 C is 796g.

5. The standard enthalpy of solution of a substance is the enthalpy change occuring when one mole of that substance is dissolved in a large volume of water. How much of a temperature ecrease will be produced by dissolving 1.0g of this ammunium nitrate in 50mL of water? The heat of solution of NH4NO3 is 25.8 kJ/mol.
G. Molar mass of NH4NO3 = 80g/mol
50mL water = 50g water
Head of solution of NH4NO3 = 25.8kJ/mol = 25800J/m
c = 4.18J/gC

R. Change in temperature

A. Q = mcT
Q = Heat of solution x # of mols of NH4NO3
Moles = Mass/Molar mass

S. #of moles = 1g / 80g/mol
= 0.0125 mols

Q = 0.0125 mols x 25800J/mol
=322.5J

322.5 = (50)(4.18)T
T = 1.5degrees

P. The water will increase by 1.5 degrees (the NH4NO3 will drop by 1.5 degrees).

6. The standard enthalpy of solution of NaOH is -42.7kJ/mol and the the standard enthalpy of neutralization of NaOH is -57.3kJ/mol. Calculate the temperature rise produced by dissolving 1g of NaOH in:
a) 100mL of water
Given: q=-42.7kJ/mol, m=100g, c=4.18J/g °C
Required:ΔT
the enthalpy of NaOH has to be converted to kJ per gram instead of moles:
-42.7kJ/mol=-1.07kJ/g
40g/mol

Equation: q sys=mcΔTsys

-1.0675kJ/g=100g(4.18J/g°C)ΔTsys

-1067.5J/g = ΔTsys
100g (4.18J/g°C)

-2.56°C=ΔTsys

ΔTsys= - ΔTsurr

-2.56°C= -ΔTsurr

2.56°C= ΔTsurr
Therefore, 100mL of water gains 2.6°C when 1g of NaOH is dissolved in it.

b) 100mL of 0.15M HCl
Part 1: Since all of the NaOH dissolves, use the amount of energy released from part a) (-1.0675 kJ)

Part 2: You have to add to this the change in temperature caused by the NaOH neutralizing.
use stoichiometry to determine how much NaOH actually neutralizes

NaOH + HCl = NaCl + H20
most of you probably know this but 0.15M HCl means that there is a concentration of HCl of 0.15 mol/L in the solution,
therefore, there is also a concentration of 0.15mol/L of NaOH to be completely neutralized
Find the Number of moles of NaOH
0.15mol/L x 0.1L = 0.0015 mol

Determine the amount of energy released by neutralization
(0.0015mol)(-57.3 kJ/mol) = -0.8595kJ

the Total amount of energy given off is the two energies added together
Qsys = (-1.0675 kJ) + (-0.8595kJ)
Qsys = -1.9265 kJ

-1926.5 J = (100g) (4.18J/g°C) ΔT
-1926.5 J = ΔT
100g(4.18J/g°C)
-4.608°C = ΔT
Therefore the change in temperature when 1g of NaOH is neutralized with 0.15M HCl is -5°C.
(I know this is not the right answer as far as the sheet goes but Leslie confirmed this is the right answer)

c) 100mL of 0.15M H2SO4

Same idea as b)
Stoichiometry
H2SO4 + 2 NaOH = Na2SO4 + 2 H2O
Therefore there is 2x as much NaOH required as H2SO4
0.15MNaOH x 2 = 0.3M NaOH
1.0g NaOH = number of moles
40g/mol
0.025mol = number of moles
0.3mol/L x 0.1 L = 0.03 mol of NaOH neutralized (theoretically)
therefore all of the NaOH is Neutralized
0.025mol x -57.3kJ/mol = -1.4325kJ

Use energy released from a)

Qtotal = -1.067kJ + (-1.4325kJ)
Qtotal = -2.4995kJ

- 2499.5 J = (100g) (4.18 J/g°C) ΔT
- 2499.5 J = ΔT
100g(4.18 J/g°C)
-5.98°C = ΔT

Therefore the change in temperature when 1g of NaOH is neutralized with 0.15M H2SO4 in 100mL of water is -6°C.

7. A man is immersed in a tub containing 60.0L of Water. The heat of his body raises the temperature of the water from 30.0°C to 31.5°C in one hour: a) At what rate is the man giving off heat? b) How many kJ would he give off in a day?

a)
Given:
ΔT = 1.5°C
Mass of H2O = 60 L = 60 kg = 60000 gNeeded:
Q of H2O per Hour
Analysis:
Q = (mass)*(change in temperature)*(specific heat capacity)

Q(H2O) = (60000 g)(1.5°C)(4.18 J/g°C)(1/h)
= 376200 J/h
= 376.2 kJ/h

Since the change in T (1.5°C) occurred in a period of an hour, 376.2 kJ is therefore the amount of heat emitted in an hour by this man.

b)

Given:
Q(H2O)/Hour = 376.2 kJ /h
t = 1 day = 24 hours
Needed:
Q(H2O)/day
Analysis:
Q/h * t = Q/day

376.2 kJ/h * 24 hours/1 day = 9028.8 kJ/day

Since the amount of heat emitted in an hour is 376.2 kJ, 9028.8 kJ is therefore the amount of heat emitted by this man in a day.